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Grace Boyd 2F
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Joined: Thu Jul 27, 2017 3:01 am


Postby Grace Boyd 2F » Mon Mar 05, 2018 7:42 pm

Hi, I am really confused by this question. For the first-order reaction A-->3B+C, when concentration initial of A = 0.015mol/L, the concentration of B increases to 0.018 mol/L in 3.0 min. a) What is the rate constant for the reaction expressed as the rate of loss of A? b) How much time would be needed for the concentration of B to increase to 0.030 mol/L?

I was wondering if someone could break this question down in steps because I am really confused since it's not just the regular "plug and chug" type question. Thanks!

Tiffany Dao 1A
Posts: 32
Joined: Fri Sep 29, 2017 7:05 am

Re: 15.29

Postby Tiffany Dao 1A » Mon Mar 05, 2018 10:28 pm

part A:
You have to change everything in terms of A, so when they give you the concentration of B after 3 minutes, to get A you would do
[A]t=[A]o - [B]t(mol A/3 mol B)
[A]t=0.015 mol A/L - (0.018mol B/L) x (mol A/3 mol B)
[A]t= 0.009 mol/L
t= 3 minutes
Then you use ln[A]t= -kt + ln[A]o
part B:
you have the rate constant from the first part, so you just need to find what the final concentration of A would be by using the same method
[A]t=0.015 mol A/L - (0.030mol B/L) x (mol A/3 mol B)
[A]t= 0.005 mol/L
Then you use ln[A]t= -kt + ln[A]o
t= 6.5 minutes
However, they are asking how much more time you would need and this is the total time.
t=6.5-3 = 3.5 minutes

Sean Monji 2B
Posts: 66
Joined: Fri Sep 29, 2017 7:06 am

Re: 15.29

Postby Sean Monji 2B » Mon Mar 05, 2018 10:37 pm

a) To find the rate constant k, we must use the rate law for a first order reaction. From the integrated law, we can find that k = ln(A initial / A at t)/t
From there, the only unknown variable is At, which can be solved using stoichiometry.
A decreases by a third of B produced, thus At = A0 - B formed in t / 3.
Then you plug in to the integrated law, finding k.

b) For this part, we are solving for t when B = 0.03 concentration. Since our integrated law is in terms of A and not B, we must find what At equals when B = 0.03. We use the same equation as part a)
At = A0 - 0.03 mols B / 3 mols B.
Then we plug into the rate law t = ln(A0 / At) / k
This will be a net time of 6.5 minutes from initial to B = 0.03. Since we know 3 minutes has already passed to get At in part a), we can subtract that from the net time.
6.5 - 3 = 3.5 minutes extra.
Hope that makes sense

Grace Han 2K
Posts: 33
Joined: Tue Nov 15, 2016 3:00 am

Re: 15.29

Postby Grace Han 2K » Wed Mar 07, 2018 10:57 pm

For this question, I got a different answer and was wondering if my method is incorrect.

Part A:

rate= 1/3 d[B]/dt= 1/3[0.018 mol L^-1/ 3min]= 0.002mol L^-1 min^-1
first order rate= k[A]
.002mol L^-1 min^-1= k[0.015 mol L^-1]
k= 0.13 min^-1

The textbook answer is k=0.17 min^-1

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